Show that the following equation has a solution in the interval (-1,1)
Multi tool use
$begingroup$
Show that the following equation has a solution in the interval (-1,1)
I think its to do with intermediate value theorem but not sure what to do.
$$
frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1} = 0
$$
real-analysis
$endgroup$
add a comment |
$begingroup$
Show that the following equation has a solution in the interval (-1,1)
I think its to do with intermediate value theorem but not sure what to do.
$$
frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1} = 0
$$
real-analysis
$endgroup$
1
$begingroup$
Well, if you know that you should start with the intermediate value theorem, then maybe you should do exactly what it states. Form a function and then look at the values of the function for the two different values of $x$ ....
$endgroup$
– Matti P.
Dec 4 '18 at 11:42
add a comment |
$begingroup$
Show that the following equation has a solution in the interval (-1,1)
I think its to do with intermediate value theorem but not sure what to do.
$$
frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1} = 0
$$
real-analysis
$endgroup$
Show that the following equation has a solution in the interval (-1,1)
I think its to do with intermediate value theorem but not sure what to do.
$$
frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1} = 0
$$
real-analysis
real-analysis
edited Dec 4 '18 at 11:41
Arthur
113k7115197
113k7115197
asked Dec 4 '18 at 11:40
user607735user607735
103
103
1
$begingroup$
Well, if you know that you should start with the intermediate value theorem, then maybe you should do exactly what it states. Form a function and then look at the values of the function for the two different values of $x$ ....
$endgroup$
– Matti P.
Dec 4 '18 at 11:42
add a comment |
1
$begingroup$
Well, if you know that you should start with the intermediate value theorem, then maybe you should do exactly what it states. Form a function and then look at the values of the function for the two different values of $x$ ....
$endgroup$
– Matti P.
Dec 4 '18 at 11:42
1
1
$begingroup$
Well, if you know that you should start with the intermediate value theorem, then maybe you should do exactly what it states. Form a function and then look at the values of the function for the two different values of $x$ ....
$endgroup$
– Matti P.
Dec 4 '18 at 11:42
$begingroup$
Well, if you know that you should start with the intermediate value theorem, then maybe you should do exactly what it states. Form a function and then look at the values of the function for the two different values of $x$ ....
$endgroup$
– Matti P.
Dec 4 '18 at 11:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The limit as $x rightarrow -1^+$ (to the right of $-1$) gives you $+infty$ because of the term $frac{e^x - 0.25}{x + 1}$, you know that $e^{-1} - 0.25 > 0$ and hence the sign of $frac{e^x - 0.25}{x + 1}$ would depend on $x+1$ which is positive as $x rightarrow -1^+$. Similarly, the limit as $x rightarrow +1^-$ (to the left of $+1$) gives you $-infty$.
Imagine how the curve would look like, it has to cut the x-axis, unless your function is piecewise continuous, which is not your case.
$endgroup$
add a comment |
$begingroup$
If calculator is allowed, you can try several numbers from the interval.
Consider the function $f(x)=frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1}, xin (-1,1)$. It is continuous in its domain.
Now calculate:
$$begin{align}f(-0.9)&=2.404>0 (f(0)=1.75>0 text{ will also do})\
f(0.9)&=-3.433<0 end{align}$$
By IVT, there must be a point $cin (-0.9,0.9)$, for which $f(c)=0$.
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add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
The limit as $x rightarrow -1^+$ (to the right of $-1$) gives you $+infty$ because of the term $frac{e^x - 0.25}{x + 1}$, you know that $e^{-1} - 0.25 > 0$ and hence the sign of $frac{e^x - 0.25}{x + 1}$ would depend on $x+1$ which is positive as $x rightarrow -1^+$. Similarly, the limit as $x rightarrow +1^-$ (to the left of $+1$) gives you $-infty$.
Imagine how the curve would look like, it has to cut the x-axis, unless your function is piecewise continuous, which is not your case.
$endgroup$
add a comment |
$begingroup$
The limit as $x rightarrow -1^+$ (to the right of $-1$) gives you $+infty$ because of the term $frac{e^x - 0.25}{x + 1}$, you know that $e^{-1} - 0.25 > 0$ and hence the sign of $frac{e^x - 0.25}{x + 1}$ would depend on $x+1$ which is positive as $x rightarrow -1^+$. Similarly, the limit as $x rightarrow +1^-$ (to the left of $+1$) gives you $-infty$.
Imagine how the curve would look like, it has to cut the x-axis, unless your function is piecewise continuous, which is not your case.
$endgroup$
add a comment |
$begingroup$
The limit as $x rightarrow -1^+$ (to the right of $-1$) gives you $+infty$ because of the term $frac{e^x - 0.25}{x + 1}$, you know that $e^{-1} - 0.25 > 0$ and hence the sign of $frac{e^x - 0.25}{x + 1}$ would depend on $x+1$ which is positive as $x rightarrow -1^+$. Similarly, the limit as $x rightarrow +1^-$ (to the left of $+1$) gives you $-infty$.
Imagine how the curve would look like, it has to cut the x-axis, unless your function is piecewise continuous, which is not your case.
$endgroup$
The limit as $x rightarrow -1^+$ (to the right of $-1$) gives you $+infty$ because of the term $frac{e^x - 0.25}{x + 1}$, you know that $e^{-1} - 0.25 > 0$ and hence the sign of $frac{e^x - 0.25}{x + 1}$ would depend on $x+1$ which is positive as $x rightarrow -1^+$. Similarly, the limit as $x rightarrow +1^-$ (to the left of $+1$) gives you $-infty$.
Imagine how the curve would look like, it has to cut the x-axis, unless your function is piecewise continuous, which is not your case.
edited Dec 4 '18 at 11:49
answered Dec 4 '18 at 11:44
Ahmad BazziAhmad Bazzi
8,0212724
8,0212724
add a comment |
add a comment |
$begingroup$
If calculator is allowed, you can try several numbers from the interval.
Consider the function $f(x)=frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1}, xin (-1,1)$. It is continuous in its domain.
Now calculate:
$$begin{align}f(-0.9)&=2.404>0 (f(0)=1.75>0 text{ will also do})\
f(0.9)&=-3.433<0 end{align}$$
By IVT, there must be a point $cin (-0.9,0.9)$, for which $f(c)=0$.
$endgroup$
add a comment |
$begingroup$
If calculator is allowed, you can try several numbers from the interval.
Consider the function $f(x)=frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1}, xin (-1,1)$. It is continuous in its domain.
Now calculate:
$$begin{align}f(-0.9)&=2.404>0 (f(0)=1.75>0 text{ will also do})\
f(0.9)&=-3.433<0 end{align}$$
By IVT, there must be a point $cin (-0.9,0.9)$, for which $f(c)=0$.
$endgroup$
add a comment |
$begingroup$
If calculator is allowed, you can try several numbers from the interval.
Consider the function $f(x)=frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1}, xin (-1,1)$. It is continuous in its domain.
Now calculate:
$$begin{align}f(-0.9)&=2.404>0 (f(0)=1.75>0 text{ will also do})\
f(0.9)&=-3.433<0 end{align}$$
By IVT, there must be a point $cin (-0.9,0.9)$, for which $f(c)=0$.
$endgroup$
If calculator is allowed, you can try several numbers from the interval.
Consider the function $f(x)=frac{e^x - 2}{x - 1} + frac{e^x - 0.25}{x + 1}, xin (-1,1)$. It is continuous in its domain.
Now calculate:
$$begin{align}f(-0.9)&=2.404>0 (f(0)=1.75>0 text{ will also do})\
f(0.9)&=-3.433<0 end{align}$$
By IVT, there must be a point $cin (-0.9,0.9)$, for which $f(c)=0$.
answered Dec 4 '18 at 12:20
farruhotafarruhota
20k2738
20k2738
add a comment |
add a comment |
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Well, if you know that you should start with the intermediate value theorem, then maybe you should do exactly what it states. Form a function and then look at the values of the function for the two different values of $x$ ....
$endgroup$
– Matti P.
Dec 4 '18 at 11:42